Answer to Question #157007 in Trigonometry for Kinza Tahir

Question

 The population of mice in Canterbury High School fluctuates from a minimum of 80 mice to a maximum of 420 mice. This population can be modelled as a function of time in months from Jan. 1st, by a cosine function. At the beginning of the year (January 1st), the population is at its least (80 mice) . After 3 months the population reaches its maximum (April 1st). On Canada Day (July 1st) the population goes back down to its minimum of 80. Three months later it reaches the maximum again (Oct. 1st). (And so on.)

(a) Draw a detailed graph of this situation for 1 year.

(b) Determine the equation of this function as a cosine function and a sine function that

describes the population of the mice in CAnterbury High School.

(c) Determine the time in months (to 1 decimal place) for one year, when the population is

above 165 mice.

Answer 1

$$y=A\cos(B(x+C))+D$$

amplitude is $A$

period is $2\pi/B$

phase shift is $C$  (positive is to the left)

vertical shift is $D$

Let $x=$ the number of months.

$$A=\dfrac{420-80}{2}=170$$

$$D=80+170=250$$

$$Period=\dfrac{2\pi}{B}=6=>B=\dfrac{\pi}{3}$$

$$y(0)=170\cos\big(\dfrac{\pi}{3}(0+C)\big)+250=80$$

$$\dfrac{\pi}{3}(C)=-\pi=>C=-3$$

$$y(x)=170\cos\big(\dfrac{\pi}{3}(x-3)\big)+250$$

(a) Draw a detailed graph of this situation for 1 year.

$$y(x)=170\cos\big(\dfrac{\pi}{3}(x-3)\big)+250, 0\leq x\leq12$$

(b)

$$y(x)=170\cos\big(\dfrac{\pi}{3}(x-3)\big)+250$$

Or

$$y(x)=170\sin\big(\dfrac{\pi}{3}(x-\dfrac{3}{2})\big)+250$$

(c)

$$y>165, 0\leq x\leq12$$

$$170\cos\big(\dfrac{\pi}{3}(x-3)\big)+250>165$$

$$\cos\big(\dfrac{\pi}{3}(x-3)\big)>-\dfrac{1}{2}$$

$$-\dfrac{2\pi}{3}+2\pi n<\dfrac{\pi}{3}(x-3)<\dfrac{2\pi}{3}+2\pi n, n\in \Z$$

$$-2+6 n<x-3<2+6 n, n\in \Z$$

$$1+6 n<x<5+6 n, n\in \Z$$

$$n=0, 1<x<5$$

$$n=1, 7<x<11$$

$x\in(1, 5)\cup(7, 11)$

During 8 months of one year the population is above 165 mice.